Elevation in boiling point is a colligative property of a solution, which means it depends on the concentration of solute particles in the solution, regardless of their identity. It is calculated using the equation ΔTb = Kb * m * i, where ΔTb is the elevation in boiling point, Kb is the molal boiling point constant, m is the molality of the solution, and i is the van't Hoff factor.

In this particular question, we are given that the solution freezes at -0.1860°C, which means the freezing point depression, ΔTf, is equal to 0.1860°C. The freezing point depression is related to the molality of the solution by the equation ΔTf = Kf * m * i, where Kf is the molal freezing point constant.

Since the same solute is used for both the freezing point depression and the boiling point elevation, the van't Hoff factor, i, is the same for both equations. Therefore, we can equate the two equations and solve for ΔTb:

ΔTf = Kf * m * i
ΔTb = Kb * m * i

0.1860 = 1.860 * m * i
ΔTb = 0.5120 * m * i

Dividing the two equations, we get:

ΔTb/ΔTf = (0.5120 * m * i) / (1.860 * m * i)

The molality and the van't Hoff factor cancel out, leaving us with:

ΔTb/ΔTf = 0.5120 / 1.860

Simplifying this equation, we find:

ΔTb/ΔTf = 0.2753

Since ΔTb/ΔTf is equal to the ratio of the boiling point elevation to the freezing point depression, we can conclude that the elevation in boiling point is 0.2753 times the freezing point depression.

Given that the freezing point depression is 0.1860°C, we can calculate the elevation in boiling point:

ΔTb = 0.2753 * 0.1860°C

ΔTb ≈ 0.0512°C

Therefore, the correct answer is option 'D': 0.0512.